A067979 Triangle read by rows of incomplete convolutions of Lucas numbers L(n+1) = A000204(n+1), n>=0.

1, 3, 6, 4, 13, 17, 7, 19, 31, 38, 11, 32, 48, 69, 80, 18, 51, 79, 107, 140, 158, 29, 83, 127, 176, 220, 274, 303, 47, 134, 206, 283, 360, 432, 519, 566, 76, 217, 333, 459, 580, 706, 822, 963, 1039, 123, 351, 539, 742, 940, 1138, 1341, 1529, 1757, 1880, 199, 568, 872, 1201, 1520, 1844, 2163, 2492, 2796, 3165, 3364

Offset: 0

Author

Wolfdieter Lang, Feb 15 2002

Keywords

• nonn
• easy
• tabl

Comments

The diagonals d>=0 (d=0: main diagonal) give convolutions of Lucas numbers L(n+1) := A000204(n+1), n>=0, with those with d-shifted index: a(d+n,d) = Sum_{k=0..n} L(k+1)*L(d+n+1-k).

The diagonals give A004799(n-1), A067980-7 for d=n-m= 0..8, respectively. Row sums give A067989.

The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(x*z)*(A(z)-x*A(x*z))/(1-x), with A(x) := (1+2*x)/(1-x-x^2) (g.f. Lucas L(n+1), n>=0).

Examples

			Triangle begins:
  {1};
  {3,6};
  {4,13,17};     p(2,x) = 4+13*x+17*x^2
  {7,19,31,38};
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A067990 (triangle with rows read backwards).

Programs

Mathematica

Table[Sum[LucasL[k + 1] LucasL[n - k + 1], {k, 0, m}], {n, 0, 10}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)

PARI

for(n=0,10, for(k=0,n, print1(sum(m=0,k,(fibonacci(m+2) + fibonacci(m))*(fibonacci(n-m+2) + fibonacci(n-m))), ", "))) \\ G. C. Greubel, Dec 17 2017

Formula

a(n, m) = Sum_{k=0..m} L(k+1)*L(n-k+1), n>=m>=0, else 0.

a(n, m) = (m+1)*L(n-m+1)*F(m) + ((m+1)*L(n-m+1) + m*L(n-m))*F(m+1), n>=m>=0, with F(n) := A000045(n) (Fibonacci) and L(n) := A000032(n) (Lucas).

G.f. for diagonals d= n-m>=0: (x^d)*(L(d+1)+L(d)*x)*(1-2*x)/(1-x-x^2)^2.

a(n, m) = -(-1)^m*F(n-2*m-1) + m*L(n+2)+F(n+3), with F(-n) = (-1)^(n+1) * F(n), hence a(n, m) = -5*A067330(n, m)+2*(m+1)*L(n+2), n>=m>=0. - Ehren Metcalfe, Apr 11 2016